Rate of Change of Vector in Rotating Frame

[ccrook]

I recognize the rate of change of a vector in an inertial frame S can be related to the rate of change of the vector in a rotating frame S₀ by the equation below taken from my textbook, where Ω is the angular velocity vector. $$\Big(\frac{dQ}{dt}\Big)_{S_{0}}= \Big(\frac{dQ}{dt}\Big)_{S} + \Omega \times Q$$
However, I am unsure which frame of reference Q refers to in the cross product.

 

[TJGilb]

Define Q.

 

[ccrook]

Define Q.

Q is an arbitrary vector such as a point of a ball in space.

 

[TJGilb]

If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.

 

[ccrook]

If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.

The origins of both frames are coincident and Q can be described in either S or S₀, but the textbook doesn't clarify which frame Q is described in for the cross product.

 

[vanhees71]

All equations are always in one frame or, even less confusing, you work with invariant objects, namely the vectors themselves, and not with their components. In this case, let's take $\vec{e}_j$ as the time-independent basis vectors in an inertial reference frame and $\vec{e}_k'$ the time-dependent vectors in the rotating frame. Then there's a rotation matrix $D_{kj}$ such that
$$\vec{e}_k'=D_{kj} \vec{e}_j.$$
Now take any vector $\vec{A}$. You have
$$\frac{\mathrm{d}}{\mathrm{d} t} \vec{A}=\dot{\vec{A}}=\frac{\mathrm{d}}{\mathrm{d} t} (A_k' \vec{e}_k').$$
Now you have
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + A_k' \dot{\vec{e}}_k'. \qquad (1)$$
But now
$$\dot{\vec{e}}_k'=\dot{D}_{kj} \vec{e}_j=\dot{D}_{kj} (D^{-1})_{jl} \vec{e}_l'=\epsilon_{klm} \Omega_m' \vec{e}_l', \qquad (2)$$
where I have used that because of
$$\hat{D} \hat{D}^T=\hat{1}$$
the matrix $\dot{D}_{kj} (D^{-1})_{jl}$ is antisymmetric (prove it!). Plugging this in (2) in (1) finally gives
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + \epsilon_{klm} A_k' \Omega_m' \vec{e}_l = (\dot{A}_l'+\epsilon_{lmk} \Omega_m' A_k') \vec{e}_l'.$$
This means that for the components of $\dot{\vec{A}}$ you get
$$\mathrm{D}_t \underline{A}'=\mathrm{d}_t \underline{A}' + \underline{\Omega}' \times \underline{A}',$$
where the underlined symbols mean the component-column vectors wrt. the basis $\vec{e}_k'$.

 

[Nugatory]

The origins of both frames are coincident and Q can be described in either S or S₀, but the textbook doesn't clarify which frame Q is described in for the cross product.

Choose some fairly easy case, such as a unit vector along the x-axis in the $S_0$ frame, and try it both ways... See which interpretation of the formula gives you a sensible result.

I'm not suggesting this because it's the best way of resolving the ambiguity in the textbook, but because the exercise will go a long ways towards building your intuition around how these transformations work.